Euler Rotation Equations and invariance of vectors in different reference frames

Question

Consider two frames, one which is inertial and the other one rotating w.r.t. to the inertial frame. Say there is a rigid body having angular momentum which is same/invariant in both frames. How is it that if the angular momentum is the same in both frames at all times, its time derivatives (given by the Euler Rotation Equations) are different?

Answer 1

This is a subtlety in the definitions. Let the inertial and rotating frames be $K$ and $K'$ respectively. Consider any vector $\mathbf{A}$ other than the angular velocity. The vector $\mathbf{A}$ itself is the same vector throughout and can be decomposed with respect to either the inertial or rotating basis, yielding different components. However, when we speak of the rate of change of $\mathbf{A}$ in the inertial and rotating frames, denoted by $\dot{\mathbf{A}}|_K$ and $\dot{\mathbf{A}}|_{K'}$ respectively, we are referring to two genuinely different vectors. They are the observed rates of change in the two frames which differ precisely because one frame is rotating and the other isn't. They are two different vectors each admitting different decompositions with respect to the inertial and rotating bases. The equation $\dot{\mathbf{A}}|_K = \dot{\mathbf{A}}|_{K'} + \boldsymbol{\omega}\times\mathbf{A}$ is a vector equation which means that it holds in any basis.

To make it more precise, let the inertial and rotating bases be $(\hat{\mathbf{i}}, \hat{\mathbf{j}}, \hat{\mathbf{k}})$ and $(\hat{\mathbf{i}}', \hat{\mathbf{j}}', \hat{\mathbf{k}}')$ respectively, located at the common fixed origin of both frames. We have $$\mathbf{A} = A_1\hat{\mathbf{i}} + A_2\hat{\mathbf{j}} + A_3 \hat{\mathbf{k}} \\ = {A_1}'\hat{\mathbf{i}}' + {A_2}'\hat{\mathbf{j}}' + {A_3}' \hat{\mathbf{k}}'.$$ Taking the time derivative in the inertial frame, we have $$\dot{\mathbf{A}}|_K = \frac{\mathrm{d}\mathbf{A}}{\mathrm{d}t} = \dot{A_1}\hat{\mathbf{i}} + \dot{A_2}\hat{\mathbf{j}} + \dot{A_3} \hat{\mathbf{k}} \\ = \dot{{A_1}'}\hat{\mathbf{i}}' + \dot{{A_2}'}\hat{\mathbf{j}}' + \dot{{A_3}'} \hat{\mathbf{k}}' + {A_1}' \frac{\mathrm{d}\hat{\mathbf{i}}'}{\mathrm{d}t} + {A_2}' \frac{\mathrm{d}\hat{\mathbf{j}}'}{\mathrm{d}t} + {A_3}' \frac{\mathrm{d}\hat{\mathbf{k}}'}{\mathrm{d}t}$$ However, taking the time derivative in the rotating frame, $$\dot{\mathbf{A}}|_{K'} = \dot{{A_1}'}\hat{\mathbf{i}}' + \dot{{A_2}'}\hat{\mathbf{j}}' + \dot{{A_3}'} \hat{\mathbf{k}}'$$ which is only the first half of the last line above because $(\hat{\mathbf{i}}', \hat{\mathbf{j}}', \hat{\mathbf{k}}')$ are constant in the rotating frame. The latter half is what we attribute to $\boldsymbol{\omega} \times \mathbf{A}$.

Answer 2

$\def \b {\mathbf}$

the angular momentum components in inertial system is:

$$\b L_I=\b R\,\b L_B\tag 1$$

where $~\b R ~$ is the transformation matrix between B-system and I-system

from equation (1)

$$\b L_I^T\,\b L_I= \left(\b R\,\b L_B\right)^T\,\b R\,\b L_B=\b L_B^T\,\b L_B\tag 2$$

thus the magnitude of the the angular momentum in I-system is equal to the magnitude of the angular momentum in B-system.

the time derivative of angular momentum equation (1) is:

$$\b{\dot{L}}_I=\b{\dot{R}}\,\b L_B+\b R\,\b{\dot{L}}_B\quad, \text{with $~\b{\dot{R}}=\b R\,\b\omega^{\times}$}$$

$$\b{\dot{L}}_I=\b R\,\left(\b\omega_B\,{\times}\,\b L_B+\b{\dot{L}}_B\right)$$

now obtain the time derivative of equation (2)

$$\b{\dot{L}}_I^T\,\b L_I+\b L_I^T\,\b{\dot{L}}_I= 2\,\b L_I^T\,\b{\dot{L}}_I=2\,\b L_B^T\,\b R^T\,\b{\dot{L}}_I= 2\,\b L_B^T\,\left(\b\omega_B\,{\times}\,\b L_B+\b{\dot{L}}_B\right)\ne\,2\,\b L_B^T\,\b{\dot{L}}_B$$